The function f(x)=4x^3-18x^2+27x-7 has:-

The correct answer is Option (3) → Point of inflexion at \( x=\displaystyle\frac{3}{2} \)

The derivative is a perfect square:

    \( f'(x)=12x^2-36x+27 \)

    \( f'(x)=3(4x^2-12x+9) \)

    \( f'(x)=3(2x-3)^2 \)

f′(x) = 0 only at \( x=\frac{3}{2} \), and \( f'(x)\gt 0 \) on both sides of it. Since f′ does not change sign, \( x=\frac{3}{2} \) is neither a maximum nor a minimum.

    \( f''(x)=24x-36 \)

    \( \displaystyle f''\left(\frac{3}{2}\right)=0 \)

f″(x) < 0 for \( x\lt\frac{3}{2} \) and f″(x) > 0 for \( x\gt\frac{3}{2} \), so the concavity changes there. Hence \( x=\frac{3}{2} \) is a point of inflexion.

Why the other options are incorrect

• Option (1) Maxima at \( x=\displaystyle\frac{3}{2} \) — Incorrect. f′(x) ≥ 0 everywhere, so f never decreases and has no maximum; f′(3/2) = 0 alone does not make a maximum.

• Option (2) Minima at \( x=\displaystyle\frac{3}{2} \) — Incorrect. For a minimum f′ must change from negative to positive, but here it is positive on both sides of \( x=\frac{3}{2} \).

• Option (4) Maximum value of function is 156 — Incorrect. f is increasing on all of R, so it has no maximum value at all.